Question: Simplify the following expression and state the condition under which the simplification is valid. $n = \dfrac{-10r^3 - 120r^2 - 360r}{-5r^3 - 35r^2 - 30r}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ n = \dfrac {-10r(r^2 + 12r + 36)} {-5r(r^2 + 7r + 6)} $ $ n = \dfrac{10r}{5r} \cdot \dfrac{r^2 + 12r + 36}{r^2 + 7r + 6} $ Simplify: $ n = 2 \cdot \dfrac{r^2 + 12r + 36}{r^2 + 7r + 6}$ Since we are dividing by $r$ , we must remember that $r \neq 0$ Next factor the numerator and denominator. $ n = 2 \cdot \dfrac{(r + 6)(r + 6)}{(r + 6)(r + 1)}$ Assuming $r \neq -6$ , we can cancel the $r + 6$ $ n = 2 \cdot \dfrac{r + 6}{r + 1}$ Therefore: $ n = \dfrac{ 2(r + 6)}{ r + 1 }$, $r \neq -6$, $r \neq 0$